# Question Description

Original Question: 931. Minimum Falling Path Sum

Given a square array of integers A, we want the minimum sum of a falling path through A.

A falling path starts at any element in the first row, and chooses one element from each row. The next row's choice must be in a column that is different from the previous row's column by at most one.

### Example:​

``Input: [[1,2,3],[4,5,6],[7,8,9]]Output: 12Explanation: The possible falling paths are:- [1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9]- [2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9]- [3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]The falling path with the smallest sum is [1,4,7], so the answer is 12.``

# Solution

``class Solution {    public int minFallingPathSum(int[][] A) {        int m = A.length, n = A[0].length;        int[][] dp = new int[m][n];        for(int i = 0; i < n; i++){            dp[0][i] = A[0][i];        }        for(int i = 1; i < m; i++){            for(int j = 0; j < n; j++){                int topLeft = j-1 < 0 ? Integer.MAX_VALUE : dp[i-1][j-1];                int topMid = dp[i-1][j];                int topRight = j+1 >= n ? Integer.MAX_VALUE : dp[i-1][j+1];                dp[i][j] = A[i][j] + Math.min(topLeft, Math.min(topMid, topRight));            }        }        int smallestPath = Integer.MAX_VALUE;        for(int i = 0; i < n; i++){            smallestPath = Math.min(smallestPath, dp[m-1][i]);        }        return smallestPath;    }}``